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Let $P$ be a positive semi-definite Hermitian matrix (i.e. $P^\dagger=P$, $x^\dagger Px \geq 0$ for all $x \in \mathbb C^n$). Then the matrix can be decomposed as $P= R+iM$ where $R$ and $M$ are strictly real matrices. Clearly $R$ is symmetric and $M$ is skew-symmetric. I call $R$ the real part of $P$.

I am almost certain that the real part $R$ must be positive semi-definite as well as I cannot find a counter-example even through numerical simulation. Is this assumption correct? Is there a way to prove the positivity of $R$ over $\mathbb C^n$ or does somebody have a counter-example?

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  • $\begingroup$ It is more common to use the Hermitian "real part" $(A+A^\dagger)/2$ $\endgroup$
    – Ben Grossmann
    Jun 20, 2017 at 20:56
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    $\begingroup$ @Omnomnomnom $P$ is already Hermitian so $(P + P^\dagger)/2 = (P + P)/2=P$ and $P$ may still have complex entries. My work requires matrices that have strictly real entries. $\endgroup$
    – gene
    Jun 20, 2017 at 21:11

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The real part of $P$ is $(P + \overline{P})/2$. If $P$ is positive semidefinite, so is $\overline{P}$, because $$x^* \overline{P} x = \overline{ \overline{x}^* P \overline{x}} \ge 0$$ And a convex combination of positive semidefinite matrices is positive semidefine.

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  • $\begingroup$ Great, thanks. Am I also able to conclude that $\overline{ \overline{x}^* P \overline{x}}=\overline{x}^* P \overline{x}$ is real since $P$ is positive semidefinite? $\endgroup$
    – gene
    Jun 20, 2017 at 21:20
  • $\begingroup$ Yes, $v^* P v$ is real whenever $P$ is hermitian. $\endgroup$
    – Robert Israel
    Jun 20, 2017 at 21:48

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